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POArtefacts from Riemann sum in scipy.signal.convolve
 

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  1. POArtefacts from Riemann sum in scipy.signal.convolve
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    <p><strong>Short summary</strong>: How do I quickly calculate the finite convolution of two arrays?</p> <h1>Problem description</h1> <p>I am trying to obtain the finite convolution of two functions f(x), g(x) defined by</p> <p><img src="https://i.stack.imgur.com/wqGyY.png" alt="finite convolution"></p> <p>To achieve this, I have taken discrete samples of the functions and turned them into arrays of length <code>steps</code>:</p> <pre><code>xarray = [x * i / steps for i in range(steps)] farray = [f(x) for x in xarray] garray = [g(x) for x in xarray] </code></pre> <p>I then tried to calculate the convolution using the <code>scipy.signal.convolve</code> function. This function gives the same results as the algorithm <code>conv</code> suggested <a href="http://www.physics.rutgers.edu/~masud/computing/WPark_recipes_in_python.html" rel="nofollow noreferrer">here</a>. However, the results differ considerably from analytical solutions. Modifying the algorithm <code>conv</code> to use the trapezoidal rule gives the desired results.</p> <p>To illustrate this, I let</p> <pre><code>f(x) = exp(-x) g(x) = 2 * exp(-2 * x) </code></pre> <p>the results are:</p> <p><img src="https://i.imgur.com/u5Kle.png" alt="enter image description here"></p> <p>Here <code>Riemann</code> represents a simple Riemann sum, <code>trapezoidal</code> is a modified version of the Riemann algorithm to use the trapezoidal rule, <code>scipy.signal.convolve</code> is the scipy function and <code>analytical</code> is the analytical convolution.</p> <p>Now let <code>g(x) = x^2 * exp(-x)</code> and the results become:</p> <p><img src="https://i.stack.imgur.com/4Xntp.png" alt="enter image description here"></p> <p>Here 'ratio' is the ratio of the values obtained from scipy to the analytical values. The above demonstrates that the problem cannot be solved by renormalising the integral.</p> <h1>The question</h1> <p>Is it possible to use the speed of scipy but retain the better results of a trapezoidal rule or do I have to write a C extension to achieve the desired results?</p> <h1>An example</h1> <p>Just copy and paste the code below to see the problem I am encountering. The two results can be brought to closer agreement by increasing the <code>steps</code> variable. I believe that the problem is due to artefacts from right hand Riemann sums because the integral is overestimated when it is increasing and approaches the analytical solution again as it is decreasing. </p> <p><strong>EDIT</strong>: I have now included the original algorithm <a href="http://www.physics.rutgers.edu/~masud/computing/WPark_recipes_in_python.html" rel="nofollow noreferrer">2</a> as a comparison which gives the same results as the <code>scipy.signal.convolve</code> function.</p> <pre><code>import numpy as np import scipy.signal as signal import matplotlib.pyplot as plt import math def convolveoriginal(x, y): ''' The original algorithm from http://www.physics.rutgers.edu/~masud/computing/WPark_recipes_in_python.html. ''' P, Q, N = len(x), len(y), len(x) + len(y) - 1 z = [] for k in range(N): t, lower, upper = 0, max(0, k - (Q - 1)), min(P - 1, k) for i in range(lower, upper + 1): t = t + x[i] * y[k - i] z.append(t) return np.array(z) #Modified to include conversion to numpy array def convolve(y1, y2, dx = None): ''' Compute the finite convolution of two signals of equal length. @param y1: First signal. @param y2: Second signal. @param dx: [optional] Integration step width. @note: Based on the algorithm at http://www.physics.rutgers.edu/~masud/computing/WPark_recipes_in_python.html. ''' P = len(y1) #Determine the length of the signal z = [] #Create a list of convolution values for k in range(P): t = 0 lower = max(0, k - (P - 1)) upper = min(P - 1, k) for i in range(lower, upper): t += (y1[i] * y2[k - i] + y1[i + 1] * y2[k - (i + 1)]) / 2 z.append(t) z = np.array(z) #Convert to a numpy array if dx != None: #Is a step width specified? z *= dx return z steps = 50 #Number of integration steps maxtime = 5 #Maximum time dt = float(maxtime) / steps #Obtain the width of a time step time = [dt * i for i in range (steps)] #Create an array of times exp1 = [math.exp(-t) for t in time] #Create an array of function values exp2 = [2 * math.exp(-2 * t) for t in time] #Calculate the analytical expression analytical = [2 * math.exp(-2 * t) * (-1 + math.exp(t)) for t in time] #Calculate the trapezoidal convolution trapezoidal = convolve(exp1, exp2, dt) #Calculate the scipy convolution sci = signal.convolve(exp1, exp2, mode = 'full') #Slice the first half to obtain the causal convolution and multiply by dt #to account for the step width sci = sci[0:steps] * dt #Calculate the convolution using the original Riemann sum algorithm riemann = convolveoriginal(exp1, exp2) riemann = riemann[0:steps] * dt #Plot plt.plot(time, analytical, label = 'analytical') plt.plot(time, trapezoidal, 'o', label = 'trapezoidal') plt.plot(time, riemann, 'o', label = 'Riemann') plt.plot(time, sci, '.', label = 'scipy.signal.convolve') plt.legend() plt.show() </code></pre> <p>Thank you for your time!</p>
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    1. COit might be helpful if you provide a [complete minimal example](http://sscce.org/) that reproduces the problem, to exclude trivial errors such as an integer division is used where true division must be used.
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    2. COif you shift `sci` array to the right (by one step) and normalize it then the solutions [look similar](http://i403.photobucket.com/albums/pp111/uber_ulrich/convolve.png): [`sci = np.r_[0,sci[:steps-1]]*0.86`](https://gist.github.com/ac45fb5d117d8ecb66a3). It seems there are different definitions of what `convolve()` means (fft, circular), see discussion in [Convolution computations in Numpy/Scipy](http://stackoverflow.com/q/6855169/4279).
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    3. COThank you for the references to the convolution thread. The right shift looks like an interesting way of fixing the problem. Can you tell me how you obtained the normalisation factor of 0.86? I have included the original Riemann sum algorithm to illustrate why I believe this is a numerical artefact rather than a different definition of what convolution means.
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